Physics - Capacitor Question with Solution | TestHub

The given circuit can be redrawn as:

Charge on $4\mathrm{ }\mu F$ capacitor:
$Q=\left(\frac{4\times 6}{4+6}\right)\times 10$
$Q=24\mathrm{ }\mathrm{\mu C}$

in series combination charge remains same.

so,

$Q_{4\mu F}=Q_{6\mu F}=24 \mu C$